Question

Cho a > 0, b> 0 và a+b = 3. Chứng minh: a² +b² ≥ \(\frac{9}{2}\)

Answer 1

Ta có: \(a+b=3\Leftrightarrow\left(a+b\right)^2=9\)

\(a^2+b^2\ge\frac{9}{2}\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow a^2+b^2\ge\frac{9}{2}\)

Answer 2

\(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{3^2}{2}=\frac{9}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=\frac{3}{2}\)